Sodium tungstate is Na₂WO4. Therefore, the formula for gold(III) tungstate is

How many atoms of gold are in gold nugget with a mass of 4.25 g

Feliz [49]

Number of Atoms in Gold for given mass can be calculated using following formula,

# of Moles = Number of Atoms / 6.022 × 10²³
Or,
Number of Atoms = Moles × 6.022 × 10²³ ------- (1)

Calculating Moles,
As,
Moles = Mass / M.mass
So,
Moles = 4.25 g / 196.96 g/mol

Moles = 0.0215

Putting value of mole in eq.1,

Number of Atoms = 0.0215 × 6.022 × 10²³

Number of Atoms = 1.299 × 10²²

Result:
4.25 g of Gold Nugget contains 1.299 × 10²² Atoms.

3 0

3 years ago

Read 2 more answers

The vapor pressure of water at 65oC is 187.54 mmHg. What is the vapor pressure of a ethylene glycol (CH2(OH)CH2(OH)) solution ma

Pavlova-9 [17]

Answer:

173.83 mmHg is the vapor pressure of a ethylene glycol solution.

Explanation:

Vapor pressure of water at 65 °C= $p_o= 187.54 mmHg$

Vapor pressure of the solution at 65 °C= $p_s$

The relative lowering of vapor pressure of solution in which non volatile solute is dissolved is equal to mole fraction of solute in the solution.

Mass of ethylene glycol = 22.37 g

Mass of water in a solution = 82.21 g

Moles of water= $n_1=\frac{82.21 g}{18 g/mol}=4.5672 mol$

Moles of ethylene glycol= $n_2=\frac{22.37 g}{62.07 g/mol}=0.3603 mol$

$\frac{p_o-p_s}{p_o}=\frac{n_2}{n_1+n_2}$

$\frac{187.54 mmHg-p_s}{187.54 mmHg}=\frac{0.3603 mol}{0.3603 mol+4.5672 mol}$

$p_s=173.83 mmHg$

173.83 mmHg is the vapor pressure of a ethylene glycol solution.

6 0

3 years ago

4) What volume will the gas in the balloon at right occupy at 250k?<br><br> balloon: 4.3L 350K

swat32

Answer:

2.87 liter.

Explanation:

Given:

Initially volume of balloon = 4.3 liter

Initially temperature of balloon = 350 K

Question asked:

What volume will the gas in the balloon occupy at 250 K ?

Solution:

By using:

$Pv =nRT$

Assuming pressure as constant,

V∝ T

Now, let K is the constant.

V = KT

Let initial volume of balloon , $V_{1}$ = 4.3 liter

1000 liter = 1 meter cube

1 liter = $\frac{1}{1000} m^{3} = 10^{-3} m^{3$

4.3 liter = $4.3\times10^{-3}=4.3\times10^{-3} m^{3}$

And initial temperature of balloon, $T_{1}$ = 350 K

Let the final volume of balloon is $V_{2}$

And as given, final temperature of balloon, $T_{2}$ is 250 K

Now, $V_{1} = KT_{1}$

$4.3\times10^{-3}=K\times350\ (equation\ 1 )$

$V_{2} = KT_{2}$

$=K\times250\ (equation 2)$

Dividing equation 1 and 2,

$\frac{4.3\times10^{-3}}{V_{2} } =\frac{K\times350}{K\times250}$

K cancelled by K.

By cross multiplication:

$350V_{2} =4.3\times10^{-3} \times250\\V_{2} =\frac{ 4.3\times10^{-3} \times250\\}{350} \\ = \frac{1075\times10^{-3}}{350} \\ =2.87\times10^{-3}m^{3}$

Now, convert it into liter with the help of calculation done above,

$2.87\times10^{-3} \times1000\\2.87\times10^{-3} \times10^{3} \\2.87\ liter$

Therefore, volume of the gas in the balloon at 250 K will be 2.87 liter.

4 0

3 years ago

Which of the following below is the complete ionic equation of the reaction: Calcium nitrate and sodium sulfide solutions react

3241004551 [841]

To find the net ionic equation we must first write the balanced equation for the reaction. We must bear in mind that the reagents Ca(NO3)2 and Na2S are in the aqueous state and as product we will have CaS in the solid state, since it is not soluble in water and NaNO3 in the aqueous state.

The balanced equation of the reaction will be:

$Ca(NO_3)_{2(aq)}+Na_2S_{(aq)}\rightarrow CaS_{(s)}+2NaNO_{3(aq)}$

Ca(NO3)2(aq) + → Ca(aq) + 2Na(s)NO3Now, c(aq)ompounds in the aqueous state can be written in their ionic form, so the reaction will transform into:Na2S +

$Ca_{(aq)}^{2+}+2(NO_3)_{(aq)}^-+2Na_{(aq)}^++S_{(aq)}^{2-}\operatorname{\rightarrow}CaS_{(s)}+2Na_{(aq)}^++2NO^-_{3(aq)}$

So, the answer will be option A

5 0

1 year ago

Does the sound wave above experience a change in pitch or loudness? Why?

Basile [38]

My Apoligies for not being sooner.

Answer: The closer the particles get or the further apart they get, the greater the sound's amplitude. Sound amplitude causes a sound's loudness and intensity. The bigger the amplitude is, the louder and more intense the sound. - (This is copied)

General Answer : Its a change in sound, bassically the bigger the waves are, the louder the sound is.

4 0

3 years ago